# Probability -- traffic lights (1. R=0.4, 2. R=0.5, 2x G=0.2)

1. Mar 7, 2016

There are 2 traffic lights. Bart knows the probability of stopping at the first light is 40%, stopping at the second light is 50% and the probability of stopping at no lights is 20%.
1. What is the probability of stopping at both lights?
2. What is the probability of stopping for at least one light?
My attempt:

Apparently the chance is 10% and not 20% of stopping at boh lights. I thought they were independent but the opposite is true. So I used the formula for dependent probability. Now I'm trying to figure out how I can find P(B=R | A=R)

Last edited: Mar 7, 2016
2. Mar 7, 2016

### BvU

But already earlier on you have a problem: case G G would be 0.6 * 0.5 and given is that it's only 0.2.

That tells you something about case G R !

The other given , 2nd light 0.5 / 0.5 on average, can then only be explained if case R G is a lot more likely than 0.4 * 0.5

3. Mar 7, 2016

### Staff: Mentor

They are dependent, yes.
You don't need formulas for this. Just consider all four options:
(A) stop at both
(B) stop at first but not at second
(C) stop at second but not at first
(D) both green

Those four options sum to a probability of 1, and you can set up three equations with the given numbers. That is sufficient to find all probabilities.

4. Mar 7, 2016

These are some serious cardinal mistakes I'm making. Thank you.

5. Mar 7, 2016

I understand my mistakes. But I really don't see how I can use the formula. To calculate the conditional probability P(B=R | A=R) I need the probability of P(B=R ∩ A=R) but that is what I'm looking for in the first place.

Our lecturer did this: P(X={red}) = 40%, P(Y={red}) = 50% (probability of red at the second light)
P(X=Red ∩ Y=Red) = P(X=R | Y=R)
P(X=Red ∩ Y=Red) = P(X=R) + P(Y=R) - P(X=R ∪ Y=R) = 40% + 50% - 80%
P(X=Red ∩ Y=Red) = 10%

I don't see how he got to P(X=R ∪ Y=R) = 80%

6. Mar 7, 2016

### BvU

OK, you picked up GG = 0.2 and hence GR = 0.4.
To end up with *G = 0.5 you need RG = 0.3 - that's one way.
To end up with *R = 0.5 you need RR = 0.1 - that's an even more direct way.
*G is the chance you need to stop at light #2.

Teacher finds chance that one or both are red = 100% - chance that both are green (which is given). That explains the 80%.
But I can't follow the logic in
can you ?

7. Mar 7, 2016

Oh I think I somewhat start to see what he did.
P(X=R) + P(Y=R) = 90% = The probability the first one is red (R) + the probability the second one is red (GR + RR)
-P(X=R ∪ Y=R) = The first light is red (R) or the second light is red (GR) or both are red (RR) ~ which equals the second question: the probability of at least one light being red = 80%

After writing this I don't see anymore. I really want to grasp this, but can't seem to find the logic. So my answer is no @BvU :(

8. Mar 7, 2016

### Staff: Mentor

You don't need conditional probabilities at all.
If they confuse you, why do you try to keep using them?

9. Mar 7, 2016

Because we need to learn to write it in a formula. I'll figure it out tomorrow. Will update this thread.

10. Mar 7, 2016

### Staff: Mentor

Then I suggest to start with the approach I posted and to calculate conditional probabilities afterwards (unnecessary, but...).

11. Mar 7, 2016

### Ray Vickson

Let R1 = {first is red (stop)}, R2 = {second is red (stop)}. You are given $P(R_1) = 0.4, P(R_2) = 0.5$ and $P(\text{not}(R_1 \cup R_2)) = 0.2$. Thus, $P(R_1 \cup R_2) = 1-0.2 = 0.8$. The inclusion-exclusion principle says that
$$P(R_1 \cup R_2) = P(R_1) + P(R_2) - P(R_1 \cap R_2),$$
so $P(R_1 \cap R_2) = 0.4 + 0.5 - 0.8 = 0.1$.